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JAMB UTME - Chemistry - 2019

Tambaya 1 Rahoto

The electronic configuration of element Z is 1s $^{2}$ 2s $^{2}$ 2p $^{6}$ 3s $^{2}$ 3p $^{1}$ . What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.

Z

_{3}

(SO

_{4}

)

_{2}

ZSO

_{4}

Z

_{2}

SO

_{4}

Z

_{2}

(SO

_{4}

)

_{3}

Bayanin Amsa

Z = 1s $^{2}$ 2s $^{2}$ 2p $^{6}$ 3s $^{2}$ 3p $^{1}$
$∴$ We have Z $^{3 +}$ and SO $_{4}^{2 -}$
The reaction : Z $^{3 +}$ + SO $_{4}^{2 -}$ $\to$ Z $_{2}$ (SO $_{4}$ ) $_{3}$ .

Karanta bayanin darasi akan Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

Tambaya 2 Rahoto

Which of the following could not be alkane?

C

_{4}

H

_{10}

C

_{5}

H

_{12}

C

_{7}

H

_{14}

C

_{8}

H

_{18}

Chemistry Of Carbon Compounds (Nigeria Only)

Tambaya 3 Rahoto

The heat of formation of ethene, C $_{2}$ H $_{4}$ is 50 kJmol $^{- 1}$ , and that of ethane, C $_{2}$ H $_{6}$ is -82kJmol $^{- 1}$ . Calculate the heat evolved in the process:

C $_{2}$ H $_{4}$ + H $_{2}$ $\to$ C $_{2}$ H $_{6}$

-122kJ

-132kJ

-88kJ

-60kJ

Energy Changes

Stoichiometry And Chemical Reactions

Tambaya 4 Rahoto

Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are

Ca(OH)

_{2}

and NaOH

Na

_{2}

CO

_{3}

NaCl

NaOH

Chemical Combination

Tambaya 6 Rahoto

Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?

2HCl + CaCO $_{3}$ → CaCl $_{2}$ + H $_{2}$ O + CO $_{2}$

Increasing the concentration of the acid

Increasing the surface area of CaCO

_{3}

A and B

Reducing the concentration of the acid

Kinetic Theory Of Matter And Gas Laws

Oxidation And Reduction

Tambaya 8 Rahoto

SO $_{3}$ is not directly dissolved in water in the industrial preparation of H $_{2}$ SO $_{4}$ by the contact process because

SO

_{3}

is an acidic gas

Acid is usually added to water and not water to acid

The reaction between SO

_{3}

and water is violently exothermic

SO

_{3}

does not dissolve readily in water

Acids, Bases And Salts

Tambaya 9 Rahoto

Hydrocarbons which will react with Tollen's reagent conform to the general formula

C

_{n}

H

_{2 n}

C

_{n}

H

_{2 n + 2}

C

_{n}

H

_{2 n - 2}

C

_{n}

H

_{2 n + 1}

OH

Chemistry Of Carbon Compounds (Nigeria Only)

Tambaya 10 Rahoto

A cell shorthand notation can be written as A / A $^{+}$ // B $^{2 +}$ /B. The double slash in the notation represents the

electrolyte

electrodes

salt bridge

solid - liquid interface

Electrolysis

Tambaya 14 Rahoto

The reaction $2 H_{2} S + S O_{2} \to 3 S + 2 H_{2} O$ is

not a redox reaction because there is no oxidant in the reaction

not a redox reaction because there is no reductant in the reaction

a redox reaction in which H

_{2}

S is the oxidant and SO

_{2}

is the reductant

a redox reaction in which SO

_{2}

is the oxidant and H

_{2}

S is the reductant

Tambaya 16 Rahoto

200cm $^{3}$ of 0.50mol/dm $^{3}$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

10g

25g

20g

15g

Bayanin Amsa

To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tambaya 18 Rahoto

Consider the equation below:

Cr $_{2}$ O $_{7}^{2 -}$ + 6Fe $^{2 +}$ + 14H $^{+}$ $\to$ 2Cr $^{3 +}$ + 6Fe $^{3 +}$ + 7H $_{2}$ O.

The oxidation number of chromium changes from

+5 to +3

+6 to +3

-2 to +3

+7 to +3

Bayanin Amsa

Cr $_{2}$ O $_{7}^{2 -}$ + 6Fe $^{2 +}$ + 14H $^{+}$ $\to$ 2Cr $^{3 +}$ + 6Fe $^{3 +}$ + 7H $_{2}$ O
The oxidation of Cr in Cr $_{2}$ O $_{7}^{2 -}$ :
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2 $⟹$ 2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3

Karanta bayanin darasi akan Oxidation And Reduction (JAMB)

Oxidation And Reduction

Tambaya 20 Rahoto

If acidified Potassium Dichromate(VI) (K $_{2}$ Cr $_{2}$ O $_{7}$ ) acts as oxidizing agent, color changes from

yellow to green

orange to green

yellow to red

orange to red

Oxidation And Reduction

Tambaya 21 Rahoto

How many alkoxyalkanes can be obtained from the molecular formula C $_{4}$ H $_{10}$ O?

4

2

1

3

Chemistry Of Carbon Compounds (Nigeria Only)

Acids, Bases And Salts (Nigeria Only)

Chemical Equilibra

Tambaya 22 Rahoto

When chlorine water is exposed to bright sunlight, the following products are formed

HCl

_{(a q)}

and Cl

_{2}

HOCl

_{(a q)}

and Cl

_{2}

HCl

_{(a q)}

and O

_{2}

HOCl

_{(a q)}

and O

_{2}

Chemistry Of Carbon Compounds (Ghana Only)

Tambaya 25 Rahoto

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.

[H = 1, C = 12, O = 16]

C

_{3}

H

_{6}

O

_{3}

C

_{6}

H

_{6}

O

_{3}

C

_{6}

H

_{12}

O

_{6}

CH

_{2}

O

Chemistry Of Carbon Compounds (Nigeria Only)

Tambaya 27 Rahoto

Which of the following reactions is an oxidation process?

2H

^{+}

+ 2e

^{-}

\to

H

_{2}

2O

^{2 -}

+ 4e

^{-}

\to

O

_{2}

OCl

^{-}

+ H

_{2}

O + 2e

^{-}

\to

2OH

^{-}

+ Cl

^{-}

Cl + e

^{-}

\to

Cl

^{-}

Oxidation And Reduction

Tambaya 28 Rahoto

By what amount must the temperature of 200cm $^{3}$ of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm $^{3}$ (Assume ideality)

150°C

177°C

75°C

120°C

Bayanin Amsa

Using the ideal gas law and equation:
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} \times 200 {cm}^{3}}{300 K} = \frac{2 P \times 150 {cm}^{3}}{T_{2}}$
Cross multiply:
$T_{2} = \frac{300 \times 150 \times 2 P}{200 \times P}$
$= 450 K$ or $177^{\circ} C$
Don't forget to convert to $^{\circ} C$

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tambaya 29 Rahoto

The oxidation state(s) of nitrogen in ammonium nitrite is/are

-3, +3

+5

-3, +5

-3

Bayanin Amsa

Ammonium nitrite = NH $_{4}$ NO $_{2}$
NH $_{4}^{+}$ : Let the oxidation number of Nitrogen = x
x + 4 = 1 $⟹$ x = 1 - 4
x = -3
NO $_{2}^{-}$ : x - 4 = -1
x = -1 + 4 $⟹$ x = +3.
The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.

Karanta bayanin darasi akan Oxidation And Reduction (JAMB)

Oxidation And Reduction

Tambaya 31 Rahoto

Elements X and Y have electronic configurations 1s $^{2}$ 2s $^{2}$ 2p $^{4}$ and 1s $^{2}$ 2s $^{2}$ 2p $^{6}$ 3s $^{2}$ 3p $^{1}$ respectively. When they combine, the formula of the compound formed is

XY

X

_{2}

Y

Y

_{2}

X

_{3}

X

_{2}

Y

_{3}

Tambaya 32 Rahoto

Consider the reaction: A + 2B $_{(g)} ⇌$ 2C + D $_{(g)}$ ( $Δ$ H = +ve)

What will be the effect of decrease in temperature on the reaction?

Rate of backward reaction will increase

The equilibrium constant will shift

Rates of forward and backward reactions are not affected

Rate of forward reaction will increase

Chemical Equilibra

Tambaya 35 Rahoto

The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?

[X = 63, Y = 27, 1F = 96,500C]

N12.00

N27.00

N63.00

N95.00

Bayanin Amsa

For X: X $^{2 +}$ + 2e $^{-}$ $\to$ X
2F = 63g
xF = 6g
x = $\frac{6 \times 2}{63} = \frac{4}{21} F$
$\frac{4}{21}$ F = N12.00
1F = $\frac{12}{\frac{4}{21}}$
= N63.00
1F is equivalent to N63.00.
For Y: Y $^{3 +}$ + 3e $^{-}$ $\to$ Y
3F = 27g
xF = 9g
x = $\frac{3 \times 9}{27}$
= 1F
1F = N63.00

Karanta bayanin darasi akan Chemical Combination (JAMB)

Chemical Combination

Tambaya 38 Rahoto

At 27°C, 58.5g of sodium chloride is present in 250cm $^{3}$ of a solution. The solubility of sodium chloride at this temperature is?

(molar mass of sodium chloride = 111.0gmol $^{- 1}$ )

2.0 moldm

^{- 3}

0.25 moldm

^{- 3}

1.0 moldm

^{- 3}

0.5 moldm

^{- 3}

Bayanin Amsa

Given the Mass of the salt = 58.5g
Volume = 250 cm $^{3}$ = 0.25 dm $^{3}$
Mass concentration = $\frac{M a s s}{V o l u m e}$
= $\frac{58.5}{0.25}$ = 234 gdm $^{- 3}$
Solubility (in moldm $^{- 3}$ = $\frac{234}{111}$
= 2.11 moldm $^{- 3}$
$≊$ 2.0 moldm $^{- 3}$

Karanta bayanin darasi akan Solubility Of Substances (Ghana Only) (WAEC)

Solubility Of Substances (Ghana Only)

Tambaya 39 Rahoto

The following are isoelectronic ions except

Na

^{+}

Mg

^{2 +}

Si

^{2 +}

O

^{2 +}

Bayanin Amsa

Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na $^{+}$ = 10 electrons = 2, 8
Mg $^{2 +}$ = 10 electrons = 2,8
O $^{2 -}$ = 10 electrons = 2,8
Si $^{2 +}$ = 12 electrons = 2,8,2
$⟹$ Si $^{2 +}$ is not isoelectronic with the rest.

Karanta bayanin darasi akan Acids, Bases And Salts (Nigeria Only) (WAEC)

Karanta bayanin darasi akan Atomic Structure And Bonding (JAMB)

Acids, Bases And Salts (Nigeria Only)

Atomic Structure And Bonding

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

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39

40